Define $f(x, y, z) = \sin(x) + y - z^2$. Let $\vec{a} = (-1, 1, -1)$ and $\vec{v} = \left( 0, 0, 1 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Explanation: When a directional derivative has a direction that equals $(1, 0, 0)$, $(0, 1, 0)$, or $(0, 0, 1)$, it becomes a regular partial derivative. Because $v = (0, 0, 1)$, the limit we want to find is the definition of $\dfrac{\partial f}{\partial z}$ evaluated at $(-1, 1, -1)$. Therefore: $ \lim_{h \to 0} \dfrac{f \left( -1, 1, -1 + h \right) - f(-1, 1, -1)}{h} = \dfrac{\partial f}{\partial z}(-1, 1, -1)$ $\begin{aligned} &\dfrac{\partial f}{\partial z} = -2z \\ \\ &\dfrac{\partial f}{\partial z}(-1, 1, -1) = 2 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = 2$.